Kinetic Theory

 $m_{N_2}=14g,T=27°C=300k$

To double the Vrms Tf = 4Ti = 4 * 300 = 1200 k.

$\therefore Q=n{C}_v\Delta T$

$=5*225R=1125R=1125*3.32$

= 9360 J.

$\overline{x}$ (mean free path) = $\frac{1}{\sqrt[]{2}\pi d^2{n}_v}$  

$n_v=\frac{n{N}_A}{v},$ n → No. of moles in volume v N_A ® Avogadro's Number

$\frac{n}{v}=\frac{P}{R_T}$

$\overline{x}\propto v\propto \frac{1}{\rho (density}\&\overline{x}\alpha TatconstantP)$

^{$\rho \uparrow \iff \overline{x}\downarrow |\overline{x}\uparrow \iff T\uparrow$}                           

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is  $=\frac{1}{2}{k}_{B}T$ (for Mono atomic gases)

This is a multiple choice answer as classified in NCERT Exemplar

When the gas is compressed adiabatically, the total work done on the gas increases its internal energy which in turn increases the KE of the gas molecule and hence, the collisions between molecules also increases.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (c) pV=nRT

(a) when pressure p =constant

volume is directly proportional to temperature

(b) when T= constant

from PV =constant

so the graph is rectangular hyperbola

(c) when V =constant

from pressure is directly proportional to temperature.

So the graph is straight the passes through the origin.

PV $\propto T$

PV/T=constant

So the graph hence through origin. So d is correct.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) The total energy associated with the molecule is

E= $\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}m{v}_z^2+\frac{1}{2}{I}_x{w}_x^2+\frac{1}{2}{I}_y{w}_y^2$

The number of independent terms in the above expressions is 5. So we can predict velocities of molecule by maxwell's distribution, hence the above expression obeys maxwell 's distribution .

So 2 rotational and 3 translational energies are associated with each molecule

So rotational energy at a given temperature is 2/3 of the translational KE of each molecule.

This is a multiple choice answer as classified in NCERT Exemplar

(c) According to kinetic theory, we know walls only exert perpendicular forces on molecule. They do not exert parallel force . so there is only change in translational motion.

So pV=2/3E

Where E is representing only translational part of energy.

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) due to the presence of external positive charge on face ABCD. The usual expression for pressure on the basis of kinetic energy will not be valid as ions would also experience electrostatic force. So presence of positive charge the isotropy is also lost.

This is a multiple choice answer as classified in NCERT Exemplar

(d) pV=nRT

So n= PV/RT

As number moles are fixed

$\frac{P_1{V}_1}{{RT}_1}=\frac{P_1{V}_1}{{RT}_1}$

P₂=p₁V₁ (T₂/V₂T₁)

= $\frac{p\mathrm{V\; (}\left(1.1\mathrm{T)}\right)}{1.05\mathrm{V\; (}\left(\mathrm{T)}\right)}$

= P $*\frac{1.1}{1.05}\approx 1.05P$

This is a multiple choice answer as classified in NCERT Exemplar

(b) For a function f (v), the number of molecules n=f (v) which are having speeds between v and v+dv

So f₁ (v) and f₂ (v) will obey's maxwell distribution law

This is a multiple choice answer as classified in NCERT Exemplar

(d) When number of molecules breaks then number of moles would become double.

P $\propto nT$ where n is the number of moles and T is temperature of ideal gas

As gases breaks number of moles becomes twcice of initial so n₂ =2n₁

$\frac{P_2}{P_1}=\frac{n_2{T}_2}{n_1{T}_1}=\frac{2n_1\left(3000\right)}{n_1\left(300\right)}$ = 20

p₂=20p₁

so 20 times